Explanation
Part fill by (A+B) in 1 minute$$=\dfrac{1}{60}+\dfrac{1}{40}=\dfrac{5}{120}=\dfrac{1}{24}$$
Let the tank is filled in x minute then
$$\dfrac{x}{2}\times \left(\dfrac{1}{24}+\dfrac{1}{40} \right)=1$$
$$\dfrac{x}{2}\times \dfrac{1}{15}=1$$
$$\dfrac{x}{30}=1$$
$$x=30 \ min$$
Then, pipe $$B$$ will fill it in $$(x+6)$$ hours.
Suppose first pipe alone takes $$x$$ hours to fill the tank
Then ,second and third pipe takes $$(x-5)$$ and $$(x-9)$$ hours to fill the tank.
$$\therefore \displaystyle \frac{1}{x}+\frac{1}{x-5}=\frac{1}{x-9}$$
$$\Rightarrow \displaystyle \frac{x-5+x}{x(x-5)}=\frac{1}{x-9}$$
$$\Rightarrow (2x-5)(x-9)=x(x-5)$$
$$\Rightarrow x^2-18x+45=0$$
$$\Rightarrow x^2-15x-3x+45=0$$
$$\Rightarrow x(x-15)-3(x-15)=0$$
$$\Rightarrow (x-3)(x-15)=0$$
$$x=15 hrs.$$
A,B and C fill tank in 6 hours. Then A,B and C one hours work =$$\dfrac{1}{6}$$
Part filled in two hours$$=\dfrac{2}{6}$$=$$\dfrac{1}{3}$$
Then remaining part $$=1-\dfrac{1}{2}=\dfrac{2}{3}$$
A and B 7 hours work $$=\dfrac{2}{3}$$
Then A and B one hours work $$=\dfrac{2}{21}$$
So C one hours work $$=(A+B+C)$$ one hours work $$-(A+B)$$ one hours work$$=\dfrac{1}{6}-\dfrac{2}{21}=$$$$\dfrac{1}{14}$$
So calone fill the tank in $$14$$ hours.
Suppose the can initially contains $$7x$$ and $$5x $$ litres of mixtures A and B respectively.
Quantity of A in mixture left$$=(7x-\cfrac { 7 }{ 12 } \times 9)=(7x-\cfrac{21}{4})$$ litres
Quantity of B in mixture left $$=(5x-\cfrac { 5}{ 12 } \times 9)=(5x-\cfrac{15}{4})$$ litres
Now,
$$\cfrac{7x-\cfrac{21}{4}}{5x-\cfrac{15}{4}+9}=\cfrac{7}{9}$$
$$=>\cfrac{28x-21}{20x+21}=\cfrac{7}{9}$$
$$=>252x-189=140x+147$$
$$=>x=3$$
Thus, the can contains $$21$$ litres of $$ A$$.
Suppose 125 men had food for x days.
Now, Less men, More days (Indirect Proportion)
$$125\times x=150\times 35$$
$$x=42days$$
Answer (C) 42days
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